题解 P3327 【[SDOI2015]约数个数和】
\[\begin{aligned} \sum_{i=1}^{n}\sum_{j=1}^{m}d(ij)
& = \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{x|i}\sum_{y|j}[gcd(x,y)=1]
\\ & = \sum_{x=1}^{n}\sum_{y=1}^{m}\:*\:{\left\lfloor\dfrac{n}{x}\right\rfloor}\:*\:{\left\lfloor\dfrac{m}{y}\right\rfloor}[gcd(x,y)=1]
\\ & = \sum_{x=1}^{n}\sum_{y=1}^{m}{\left\lfloor\dfrac{n}{x}\right\rfloor}\:*\:{\left\lfloor\dfrac{m}{y}\right\rfloor}\:*\:\sum_{t|gcd(x,y)}\mu(t)
\\ & = \sum_{t=1}^{min(n,m)}\mu(t)\times \sum_{x=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}\sum_{y=1}^{\left\lfloor\dfrac{m}{t}\right\rfloor}\left\lfloor\dfrac{n}{xt}\right\rfloor\:*\:\left\lfloor\dfrac{m}{xt}\right\rfloor
\\ & = \sum_{t=1}^{min(n,m)}\mu(t)\times \sum_{x=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}\left\lfloor\dfrac{\left\lfloor\dfrac{n}{x}\right\rfloor}{t}\right\rfloor\times \sum_{x=1}^{\left\lfloor\dfrac{m}{t}\right\rfloor}\left\lfloor\dfrac{\left\lfloor\dfrac{m}{x}\right\rfloor}{t}\right\rfloor
\end{aligned}
\]
\[\text{莫比乌斯反演的“套路”}
\text{设}f(x) = \sum_{i=1}^{x}{\left\lfloor\dfrac{x}{i}\right\rfloor}
\]
\[ans=\sum_{t=1}^{min(n,m)}\mu(t)\times
f(\left\lfloor\dfrac{n}{t}\right\rfloor)\times
f(\left\lfloor\dfrac{m}{t}\right\rfloor)
\]
\(\mu\)用前缀和
而f函数可以用数论分块\(O(\sqrt{N})\)
